# M.E. Irizarry-Gelpí

Physics impostor. Mathematics interloper. Husband. Father.

# Ann, Bob, and Eve 3

Ann has two ortho-normal bases at her disposal for encoding cbits into qbits:

\begin{align*} \lvert A_{0} \rangle &= \cos(\theta_{a}) \lvert 0 \rangle - \sin(\theta_{a}) \lvert 1 \rangle & \lvert A_{1} \rangle &= \sin(\theta_{a}) \lvert 0 \rangle + \cos(\theta_{a}) \lvert 1 \rangle \\ \lvert B_{0} \rangle &= \cos(\theta_{b}) \lvert 0 \rangle - \sin(\theta_{b}) \lvert 1 \rangle & \lvert B_{1} \rangle &= \sin(\theta_{b}) \lvert 0 \rangle + \cos(\theta_{b}) \lvert 1 \rangle \end{align*}

Note that for $$\theta_{a} = 0$$, the $$A$$ basis becomes

\begin{align*} \lvert A_{0} \rangle &= \lvert 0 \rangle & \lvert A_{1} \rangle &= \lvert 1 \rangle \end{align*}

and for $$\theta_{b} = \pi/4$$, the $$B$$ basis becomes

\begin{align*} \lvert B_{0} \rangle &= \lvert {-} \rangle & \lvert B_{1} \rangle &= \lvert {+} \rangle \end{align*}

Bob uses these two bases for decoding. Eve can use a third ortho-normal basis:

\begin{align*} \lvert C_{0} \rangle &= \cos(\theta_{c}) \lvert 0 \rangle - \sin(\theta_{c}) \lvert 1 \rangle & \lvert C_{1} \rangle &= \sin(\theta_{c}) \lvert 0 \rangle + \cos(\theta_{c}) \lvert 1 \rangle \end{align*}

You can check that

\begin{align*} \lvert A_{0} \rangle \langle A_{0} \rvert &= \begin{bmatrix} \cos^{2}(\theta_{a}) & -\cos(\theta_{a}) \sin(\theta_{a}) \\ -\cos(\theta_{a}) \sin(\theta_{a}) & \sin^{2}(\theta_{a}) \end{bmatrix} \\ \lvert A_{1} \rangle \langle A_{1} \rvert &= \begin{bmatrix} \sin^{2}(\theta_{a}) & \cos(\theta_{a}) \sin(\theta_{a}) \\ \cos(\theta_{a}) \sin(\theta_{a}) & \cos^{2}(\theta_{a}) \end{bmatrix} \end{align*}

Let

\begin{align*} \theta_{ab} &\equiv \theta_{a} - \theta_{b} & \theta_{ac} &\equiv \theta_{a} - \theta_{c} &\theta_{bc} &\equiv \theta_{b} - \theta_{c} \end{align*}

Then, for example:

\begin{align*} \langle C_{0} \vert A_{0} \rangle &= \cos(\theta_{c})\cos(\theta_{a}) + \sin(\theta_{c}) \sin(\theta_{a}) = \cos(\theta_{ac}) \\ \langle C_{0} \vert A_{1} \rangle &= \cos(\theta_{c})\sin(\theta_{a}) - \sin(\theta_{c}) \cos(\theta_{a}) = \sin(\theta_{ac}) \\ \langle C_{1} \vert A_{0} \rangle &= \sin(\theta_{c})\cos(\theta_{a}) - \cos(\theta_{c}) \sin(\theta_{a}) = -\sin(\theta_{ac}) \\ \langle C_{1} \vert A_{1} \rangle &= \sin(\theta_{c})\sin(\theta_{a}) + \cos(\theta_{c}) \cos(\theta_{a}) = \cos(\theta_{ac}) \end{align*}

We are going to consider three scenarios where Ann and Bob share a message and Eve eavesdrops:

1. Eve does not eavesdrops
2. Eve eavesdrops with a random choice of basis $$A$$ or $$B$$
3. Eve eavesdrops with basis $$C$$

In this post we will go over scenario 3, where Eve eavesdrops with basis $$C$$.

Ann flips a fair coin and chooses an encoding basis from $$A$$ or $$B$$. Bob flips another coin and chooses a decoding basis from $$A$$ or $$B$$. If they do not make the same basis choice, they discard that event. Eve always uses the same eavesdropping basis ($$C$$). Now there are two possible events:

1. Ann encodes with $$A$$, Eve eavesdrops with $$C$$, Bob decodes with $$A$$
2. Ann encodes with $$B$$, Eve eavesdrops with $$C$$, Bob decodes with $$B$$

Consider event 1. Ann prepares the state $$\lvert A_{0} \rangle$$. The density matrix is

\begin{equation*} \rho = \lvert A_{0} \rangle \langle A_{0} \rvert \end{equation*}

Then Eve eavesdrops with the $$C$$ basis. The probability that Eve gets $$\lvert C_{0} \rangle$$ is

\begin{equation*} p(C_{0} \vert \rho) = \langle C_{0} \vert \rho \vert C_{0} \rangle = \langle C_{0} \vert A_{0} \rangle \langle A_{0} \vert C_{0} \rangle = \cos^{2}(\theta_{ac}) \end{equation*}

After the measurement, the density matrix is given by

\begin{equation*} \rho_{0} = \lvert C_{0} \rangle \langle C_{0} \rvert \end{equation*}

The probability that Eve gets $$\lvert C_{1} \rangle$$ is

\begin{equation*} p(C_{1} \vert \rho) = \langle C_{1} \vert \rho \vert C_{1} \rangle = \langle C_{1} \vert A_{0} \rangle \langle A_{0} \vert C_{1} \rangle = \sin^{2}(\theta_{ac}) \end{equation*}

Right after the measurement, the density matrix is given by

\begin{equation*} \rho_{1} = \lvert C_{1} \rangle \langle C_{1} \rvert \end{equation*}

Thus, the density matrix that Bob receives is

\begin{equation*} \rho = \cos^{2}(\theta_{ac}) \lvert C_{0} \rangle \langle C_{0} \rvert + \sin^{2}(\theta_{ac}) \lvert C_{1} \rangle \langle C_{1} \rvert \end{equation*}

Bob then decodes with the $$A$$ basis. The probability that Bob gets $$\lvert A_{0} \rangle$$ is

\begin{equation*} p(A_{0} \vert \rho) = \langle A_{0} \vert \rho \vert A_{0} \rangle = \cos^{4}(\theta_{ac}) + \sin^{4}(\theta_{ac}) \end{equation*}

The probability that Bob gets $$\lvert A_{1} \rangle$$ is

\begin{equation*} p(A_{1} \vert \rho) = \langle A_{1} \vert \rho \vert A_{1} \rangle = 2\cos^{2}(\theta_{ac}) \sin^{2}(\theta_{ac}) \end{equation*}

Thus, the probability of success for event 1 is

\begin{equation*} p_{1} = \cos^{4}(\theta_{ac}) + \sin^{4}(\theta_{ac}) \end{equation*}

In a very similar manner, you can show that the probability of success for event 2 is

\begin{equation*} p_{2} = \cos^{4}(\theta_{bc}) + \sin^{4}(\theta_{bc}) \end{equation*}

Thus, the overall probability of success is

\begin{equation*} P = \frac{p_{1} + p_{2}}{2} = \frac{\cos^{4}(\theta_{ac}) + \sin^{4}(\theta_{ac}) + \cos^{4}(\theta_{bc}) + \sin^{4}(\theta_{bc})}{2} \end{equation*}

(Yet again, we have assumed that there is no noise.)